| Karim Belabas on Thu, 18 Jun 2009 11:28:59 +0200 |
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| Re: Elliptic curve x^3 - y^2 = p |
* cino hilliard [2009-06-18 10:59]:
> I want to find the number of solutions of the elliptic curve, x^3 - y^2 = p
>
> for various p = 7, 431, 503, etc
>
>
>
> I have been using brute force in a Pari script below testing for solutions.
>
> diffcubesq2(n,p) =
> {
> local(a,c=0,c2=0,j,k,y);
> for(j=1,n,
> for(k=1,n,
> y=j^3-k^2;
> if(y==p,
> c++;
> print(j","k","y);
> );
> );
> );
> c;
>
> }
>
>
> diffcubesq2(10000,431) outputs
>
> 8,9,431
> 11,30,431
> 20,87,431
> 30,163,431
> 36,215,431
> 138,1621,431
> 150,1837,431
>
> (03:14:10) gp > ##
> *** last result computed in 6mn, 57,969 ms.
Here's a "simpler" and better approach (still naive) for your problem:
diffcubes(n, p)=
setintersect(vector(n, x, x^3 - p), vector(n, y, y^2));
(11:15) gp > diffcubes(10000,431)
time = 10 ms.
%2 = [81, 900, 7569, 26569, 46225, 2627641, 3374569]
I trust you can work out the individual solutions (x,y) from the above data :-)
For each given p, you can certainly work out necessary congruence conditions
and restrict to arithmetic progressions for linear speedups.
> My Pari code misses the last two solutions. It would have
>
> taken way too much time to get to y = 243836 anyway.
(11:19) gp > diffcubes(243836, 431)
time = 130 ms.
%3 = [81, 900, 7569, 26569, 46225, 2627641, 3374569, 190108944, 59455994896]
> I tried using the Magma applet to compute the elliptic curve.
> This gets all solutions in a fraction of the time.
[...]
> E := EllipticCurve([0, -7]);
> Q, reps := IntegralPoints(E);
This is a much more sophisticated algorithm, involving computing the
full Mordell-Weil group, then transcendence methods (linear forms in
elliptic logarithms + de Weger's reduction).
Cheers,
K.B.
--
Karim Belabas, IMB (UMR 5251) Tel: (+33) (0)5 40 00 26 17
Universite Bordeaux 1 Fax: (+33) (0)5 40 00 69 50
351, cours de la Liberation http://www.math.u-bordeaux1.fr/~belabas/
F-33405 Talence (France) http://pari.math.u-bordeaux1.fr/ [PARI/GP]
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